What is the sum from n=n=1 to infinity of 1/(2^n) ?

The sum from n=n=1 to infinity of 1/(2^n) is converges

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sum from n=n=1 to infinity of 1/(2^n)

n = n = 1 \sum \infty \frac{1}{2 ^{n}}

converges

Solution steps

n = n = 1 \sum \infty \frac{1}{2 ^{n}}

Apply Series Ratio Test:

converges

= converges

k = 0 \sum \infty (\frac{3}{4})^{k}

n = 1 \sum \infty (1 - \frac{2}{n})^{3 n}

n = 6 \sum \infty \frac{2}{n - 5}

n = 0 \sum \infty \frac{1}{e ^{- n}}

n = 0 \sum \infty \frac{1}{1 + 3 n ^{3}}

What is the sum from n=n=1 to infinity of 1/(2^n) ?
The sum from n=n=1 to infinity of 1/(2^n) is converges